Given,

$\underset{x\to 0}{\lim }\frac{2^{2x}-2^{x+1}+2-\cos 2x}{x^2}$

$=\underset{x\to 0}{\lim }\frac{2^{2x}-2\times 2^x+1+1-\cos 2x}{x^2}$

$=\underset{x\to 0}{\lim }\frac{{\left(2^x-1\right)}^2+1-\cos 2x}{x^2}$

$=\underset{x\to 0}{\lim }{\left(\frac{2^x-1}{x}\right)}^2+4\times \underset{x\to 0}{\lim }\frac{1-\cos 2x}{4x^2}$

$=\underset{x\to 0}{\lim }{\left(\frac{2^x-1}{x}\right)}^2+4\times \underset{x\to 0}{\lim }\frac{1-\cos 2x}{{\left(2x\right)}^2}$

Now using the formula $\underset{x\to 0}{\lim }\frac{a^x-1}{x}={\log }_{e}a$ and $\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{1}{2}$ we get,

$\underset{x\to 0}{\lim }{\left(\frac{2^x-1}{x}\right)}^2+4\times \underset{x\to 0}{\lim }\frac{1-\cos 2x}{{\left(2x\right)}^2}$

$={\left({\log }_{e}2\right)}^2+4\times \frac{1}{2}={\left({\log }_{e}2\right)}^2+2$